求$\iint_R (x^2 + y^2) dA$,其中 $R$ 是由 $y = x$ 和 $y = x^2$ 所围成的区域
时间: 2023-09-05 10:11:55 浏览: 148
首先,我们可以画出该区域 $R$ 和其边界:
![image.png](https://cdn.nlark.com/yuque/0/2021/png/2362012/1629881011409-2f1f7ebd-8316-4c9a-9b3c-2f1d3c09d1f0.png#align=left&display=inline&height=156&margin=%5Bobject%20Object%5D&name=image.png&originHeight=156&originWidth=268&size=4746&status=done&style=none&width=268)
接下来,我们可以通过将积分区域投影到 $x$ 轴和 $y$ 轴上,确定积分的范围:
$$
x^2 \leq y \leq x, \quad 0 \leq x \leq 1
$$
$$
y \leq x^2, \quad 1 \leq x \leq \sqrt{2}
$$
因此,原式可以表示为以下积分形式:
$$
\begin{aligned}
\iint_R (x^2 + y^2) dA &= \int_0^1 \int_{x^2}^x (x^2 + y^2) dy dx + \int_1^{\sqrt{2}} \int_{y^{1/2}}^{y^{1/2}} (x^2 + y^2) dx dy \\
&= \int_0^1 \left[xy + \frac{y^2}{2}\right]_{x^2}^x dx + \int_1^{\sqrt{2}} \left[\frac{x^3}{3} + xy\right]_{y^{1/2}}^{y} dy \\
&= \int_0^1 \left(\frac{3}{2}x^4 - \frac{1}{2}x^6\right)dx + \int_1^{\sqrt{2}} \left(\frac{1}{3}y^3 + \frac{1}{2}y^2 - y^{3/2}\right)dy \\
&= \left[\frac{3}{10}x^5 - \frac{1}{14}x^7\right]_0^1 + \left[\frac{1}{12}y^4 + \frac{1}{6}y^3 - \frac{2}{5}y^{5/2}\right]_1^{\sqrt{2}} \\
&= \frac{3}{10} - \frac{1}{14} + \frac{1}{12}(\sqrt{2})^4 + \frac{1}{6}(\sqrt{2})^3 - \frac{2}{5}(\sqrt{2})^{5/2} - \frac{1}{12} - \frac{1}{6} + \frac{2}{5} \\
&= \frac{13\sqrt{2}}{35} - \frac{13}{210}
\end{aligned}
$$
因此,$\iint_R (x^2 + y^2) dA = \frac{13\sqrt{2}}{35} - \frac{13}{210}$。
阅读全文